# How do you solve 15x-10y=-5 and 15y=5+10x using substitution?

Jun 27, 2018

$x = - \frac{1}{5} , y = \frac{1}{5}$

#### Explanation:

Since every term in both equations is divisible by $5$, let's divide every term by this. We now have the equations

$3 x - 2 y = - 1$

$3 y = 1 + 2 x$

In the second equation, let's divide all terms by $3$ to solve for $y$. We get

$\textcolor{b l u e}{y = \frac{2}{3} x + \frac{1}{3}}$

This is where the substitution comes in...let's plug this value for $y$ into the first equation.

$3 x - 2 \textcolor{b l u e}{\left(\frac{2}{3} x + \frac{1}{3}\right)} = - 1$

Distributing the $- 2$, we get

$3 x - \frac{4}{3} x - \frac{2}{3} = - 1$

Let's make all terms have a common denominator:

$\frac{9}{3} x - \frac{4}{3} x - \frac{2}{3} = - \frac{3}{3}$

Now, we can simplify this further:

$\frac{5}{3} x - \frac{2}{3} = - \frac{3}{3}$

$\implies \frac{5}{3} x = - \frac{1}{3}$

$\implies 5 x = - 1$

$\implies \textcolor{\mathrm{da} r k v i o \le t}{x = - \frac{1}{5}}$

Let's solve for $y$ now. We can plug this $x$ value into our blue equation:

$y = \frac{2}{3} \textcolor{\mathrm{da} r k v i o \le t}{\left(- \frac{1}{5}\right)} + \frac{1}{3}$

$\implies y = - \frac{2}{15} + \frac{1}{3}$

$\implies y = - \frac{2}{15} + \frac{5}{15} = \frac{3}{15} = \frac{1}{5}$

$x = - \frac{1}{5} , y = \frac{1}{5}$

Hope this helps!